Let $f(x) = -9x^{2}+9x+9$. Where does this function intersect the x-axis (i.e. what are the roots or zeroes of $f(x)$ )?
Answer: The function intersects the x-axis when $f(x) = 0$ , so you need to solve the equation: $-9x^{2}+9x+9 = 0$ Use the quadratic formula to solve $ax^2 + bx + c = 0$ $x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ $a = -9, b = 9, c = 9$ $ x = \dfrac{-9 \pm \sqrt{9^{2} - 4 \cdot -9 \cdot 9}}{2 \cdot -9}$ $ x = \dfrac{-9 \pm \sqrt{405}}{-18}$ $ x = \dfrac{-9 \pm 9\sqrt{5}}{-18}$ $x =\dfrac{-1 \pm \sqrt{5}}{-2}$